Combinations Without Repetition Calculator

Compute C(n, r) = n! / (r!(n − r)!), the number of ways to choose r items from n distinct items when order does not matter and no item is reused.

Quick Facts

Formula
C(n, r) = n! / (r! (n − r)!)
Counts unordered selections with no repeats; equals the binomial coefficient "n choose r".

Your Results

Calculated
Combinations C(n, r)
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Ways to choose r from n
Permutations P(n, r)
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If order mattered: n!/(n−r)!
Orderings per combination
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r! arrangements of each selection

Ready

Enter n and r, then press Calculate.

Combinations without repetition, explained

A combination without repetition answers a simple question: in how many ways can you choose r items from a set of n distinct items when the order of your choice does not matter and no item can be picked more than once? Because order is ignored, choosing {A, B, C} is the same outcome as choosing {C, A, B} — they are counted once, not six times. This quantity is written C(n, r), nCr, or "n choose r," and it is called the binomial coefficient.

The formula

The number of combinations without repetition is:

C(n, r) = n! / ( r! · (n − r)! )

Here "!" is the factorial: n! = n × (n − 1) × (n − 2) × … × 2 × 1, and by convention 0! = 1. The r! in the denominator divides out the orderings within your chosen group (each set of r items can be arranged in r! ways), and (n − r)! divides out the orderings of the items you left behind. What remains is the count of distinct unordered selections.

A more computation-friendly form avoids huge factorials by cancelling (n − r)! from the top and bottom:

C(n, r) = [ n · (n − 1) · … · (n − r + 1) ] / r!

This calculator uses that multiplicative form so it stays accurate for larger n and r than the naive factorial approach can handle.

A worked example: the lottery

A "6/49" lottery draws 6 numbers from 49, and the order of the draw does not matter — so it is a combination without repetition. C(49, 6) = 49! / (6! · 43!) = (49·48·47·46·45·44) / (6·5·4·3·2·1) = 10,068,347,520 / 720 = 13,983,816. That is why a single ticket has roughly a 1-in-14-million chance of matching all six numbers.

Combinations versus permutations

If order did matter, you would count permutations instead: P(n, r) = n! / (n − r)!. Every combination of r items can be arranged in r! different orders, so the two are linked by C(n, r) = P(n, r) / r!. Use combinations for a committee, a hand of cards, a pizza's toppings, or lottery numbers; use permutations for a ranked podium, a PIN, or a race finish where sequence matters.

Handy reference values

  • Symmetry: C(n, r) = C(n, n − r). Choosing which r to include is the same as choosing which n − r to exclude — so C(10, 3) = C(10, 7) = 120.
  • Edges: C(n, 0) = 1 (one way to choose nothing) and C(n, n) = 1 (one way to choose everything).
  • C(52, 5) = 2,598,960 — the number of distinct 5-card poker hands.
  • C(52, 13) = 635,013,559,600 — the number of possible 13-card bridge hands.
  • Row sums: the values C(n, 0), C(n, 1), …, C(n, n) form a row of Pascal's Triangle and add up to 2n.

Frequently Asked Questions

What is the formula for combinations without repetition?
C(n, r) = n! / (r! · (n − r)!), where n is the number of distinct items available and r is the number you choose. It counts unordered selections in which each item appears at most once. For instance, C(5, 2) = 120 / (2 × 6) = 10.
What is the difference between a combination and a permutation?
A permutation counts ordered arrangements, so ABC and CBA are counted separately. A combination counts unordered selections, so ABC and CBA are the same choice. They are related by C(n, r) = P(n, r) / r!, because each combination of r items can be ordered in r! ways.
What happens if r is greater than n?
C(n, r) = 0 whenever r > n. You cannot choose more distinct items than exist in the set, so there are zero valid selections. This calculator flags that case rather than returning a number.
Why is 0! equal to 1?
It is defined that way so the formula stays consistent. There is exactly one way to arrange an empty set (do nothing), and setting 0! = 1 makes C(n, 0) = 1 and C(n, n) = 1 come out correctly, which matches the fact that there is one way to choose none and one way to choose all.

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